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From: "Dr. Sam Lomonaco"Date: Sun, 17 Oct 1999 00:06:49 -0400 (EDT) To: ???? Subject: Re: CMSC 442 Homework #4 Cc: lomonaco@umbc.edu ????: > >Well, I *thought* I understood everything we'd done in class, but I am >having a bit of trouble with problem 1, and *lots* of trouble with >problem 3. > u lies in the ideal (x^2 + x + 1) iff there is an element a in the ring R_3 such that u=a*(x^2 + x + 1). The elements of R_3 can all be represented in terms of polynomials of degree < 3, since x^3 = 1. Therefore, a = (a_2)x^2 + (a_1)x + a_0, where the a_k's are either zero or one. This using the relation x^3 = 1, you can with a little calculation show that a*(x^2 + x + 1) = (a_2 + a_1 + a_0)*(x^2 + x + 1) Thus, (x^2 + x + 1) = { 0, x^2 + x + 1 } > >I'm not quite sure how to compute the orders for ksi^i i = 0,1,....,62. > The order of a is the smallest positive integer k such that a^k = 1 An element ksi^i is primitive if it has maximal order (2^6)-1. In other words, if the powers of (ksi^i)^k, k=0..(2^6)-2 are all the non-zero elements of GF(2^6). ..... Regards, Dr. L .................................................................. Samuel J. Lomonaco | Univ. of Maryland Baltimore County (UMBC) | .................................................................| Email: | 410-455-3500 (Sec) | 410-455-2338 (O) | Lomonaco@UMBC.Edu | 410-455-3969(FAX) | | .................................................................| World Wide Web Page: http://www.cs.umbc.edu/~lomonaco/ | ..................................................................